The center-radius form of the circle equation is given by :
#(x-h)^2+(y-k)^2=r^2#
where #h and k# are the coordinates of the center of the circle, and #r# is the radius.
Given the equation of the circle is :
#x^2+y^2-4x-6y+9=0#
Rewrite this equation into the center-radius form,
# x^2+y^2-4x-6y+9=0#
#=> x^2-4x+y^2-6y=-9#
#=> (x^2-4x+color(red)4)+(y^2-6y+color(red)9)=-9+color(red)4+color(red)9#
#=> (x-2)^2+(y-3)^2=4#
#=> (x-2)^2+(y-3)^2=2^2#
So the center of the circle is at #(2,3)#, and the radius #=2#
As shown in the figure, #O# is the center of the circle, #Q# is the point of tangency, and #OQ# is the radius.
#OPQ# forms a right triangle.
By Pythagorean theorem : #PO^2=PQ^2+OQ^2#
Use the distance formula to find #PO#,
#PO=sqrt((5-2)^2+(7-3)^2)=sqrt(9+16)=sqrt25=5#
#=> PQ^2=PO^2-OQ^2=5^2-2^2=25-4=21#
#=> PQ=+-sqrt21#,
As #PQ# can not be negative, #=> PQ=sqrt21# unit
Hence, length of tangent #=PQ=sqrt21# units
Sol. 2) : recall that length of tangent from point #P(x_1,y_1)# to the circle #x^2+y^2+2gx+2fy+c=0# is :
#sqrt(x_1^2+y_1^2+2gx_1+2fy_1+c)#,
#=> PQ=sqrt(5^2+7^2-(4xx5)-(6xx7)+9)=sqrt21# units
